Howdy all! I've ran into a slight issue and I hope you guys can clarify where I went wrong.

 

Basically there I was today sitting at some boring meeting barely listening and out of boredom was trying to derive formulas for ballistic projectile trajectory. Unfortunately, when meeting finally finished and I went into wiki and compared what I have to what the actual formulas are mine were incorrect and I've ran some numbers and I definitely get wrong results. It's all very embarrassing, but I've checked my scribbles and all seems reasonable.

 

Anyways, I went like this:

 

At point 1 on the surface level we have a projectile launched with a velocity [latex]v_1[/latex] at an angle [latex]\alpha[/latex].

 

At point 2 the projectile has reached maximum height and about to start falling back down.

 

By energy conservation we have [latex]E_{k1} + E_{p1} = E_{k2} + E_{p2}[/latex]. Obviously, at point 1 projectile being at surface level the potential energy is 0, then:

 

[latex]\frac{mv_1^2}{2} = \frac{mv_2^2}{2} + mgh[/latex]

 

Dividing both sides by m and multiplying by 2 we get [latex]v_1^2 = v_2^2 + 2gh[/latex], but since the vertical component of the original [latex]v_1[/latex] vector is now 0, [latex]v_2[/latex] is essentially the horizontal component of the starting velocity or [latex]v_1*cos(\alpha)[/latex]. Then,

 

[latex]v_1^2 = v_1^2*cos^2(\alpha)+2gh[/latex]

 

Simplifying that I got:

 

[latex]h = \frac{v_1^2(1-cos^2(\alpha))}{2g}[/latex], or [latex]h = \frac{v_1^2sin^2(\alpha)}{2g}[/latex]

 

After that I went to derive the range of projectile. Since two parts of trajectory, namely, surface to highest point and highest point back to surface are essentially mirror images of one another I got:

 

[latex]R = v_{1h}*2t [/latex], where [latex]v_{1h}[/latex] is the horizontal component of the starting velocity and [latex]t[/latex] is the time required for projectile to get to the highest point of trajectory.

 

At point 2 vertical velocity is 0, then:

 

[latex]v_{1v} - gt^2/2 = 0[/latex]

 

simplifying for t,

 

[latex]t = \sqrt{\frac{2v_{1v}}{g}}[/latex], but since [latex]v_{1v} = v_1*sin(\alpha)[/latex]

 

[latex]t = \sqrt{\frac{2v_{1}*sin(\alpha)}{g}}[/latex]

 

And then range becomes:

 

[latex]R=v_{1h}*2*\sqrt{\frac{2v_{1}*sin(\alpha)}{g}} = 2v_{1}cos(\alpha)\sqrt{\frac{2v_{1}*sin(\alpha)}{g}}[/latex]

 

And this is all wrong. I've definitely done some very stupid mistake but for the life of me can't find where exactly. Any ideas?

Edited October 14, 201510 yr by pavelcherepan

 

[latex]R = v_{1h}*2t [/latex], where [latex]v_{1h}[/latex] is the horizontal component of the starting velocity and [latex]t[/latex] is the time required for projectile to get to the highest point of trajectory.

 

At point 2 vertical velocity is 0, then:

 

[latex]v_{1v} - gt^2/2 = 0[/latex]

 

 

Why is this set equal to zero? And do the units work?

 

At point 2 vertical velocity is 0, then:

 

 

 

This is incorrect.

 

t is not squared in the formula final velocity = initial velocity + acceleration times time (V = U +ft)

 

Be careful with your signs as well is g positive or negative?

 

the correct expression for time to vertex is

 

t =( v₁sin(a))/2

Edited October 14, 201510 yr by studiot

Oy vey! I knew it was some very stupid mistake. [latex]\frac{gt^2}{2}[/latex] is the distance traveled by free-falling body, not sure why I used it here.

 

Thanks for the help guys!

Oy vey! I knew it was some very stupid mistake. [latex]\frac{gt^2}{2}[/latex] is the distance traveled by free-falling body, not sure why I used it here.

 

Thanks for the help guys!

 

it looked like a bad mixture of v = u+at and s=ut +1/2 at^2. As per both above posters; whenever you reach an impasse like this check your dimension a/o units whichever you feel most comfortable with

 

your equation was

ms^-1 = ms^-2 s² or [length] [time]^-1 = [length] [time]^-2 [time]²

which are clearly wrong